This is a question from the book Introduction to Set Theory (Hrbacek and Jech), chapter 5, question 2.6.
(Show that) The cardinality of the set of all discontinuous functions is $2^{2^{\aleph_0}}$. [Hint: Using exercise 2.5, show that $|\mathbb{R}^\mathbb{R}-C|=2^{2^{\aleph_0}}$ whenever $|C|\leq 2^{\aleph_0}$.]
Exercise 2.5, as referred to in the hint, provides the following result:
For $n>0$, $n \cdot 2^{2^{\aleph_0}} = \aleph_0 \cdot 2^{2^{\aleph_0}} = 2^{\aleph_0} \cdot 2^{2^{\aleph_0}} = 2^{2^{\aleph_0}} \cdot 2^{2^{\aleph_0}} = (2^{2^{\aleph_0}})^n = (2^{2^{\aleph_0}})^{\aleph_0} = (2^{2^{\aleph_0}})^{2^{\aleph_0}}=2^{2^{\aleph_0}}$.
The only way I could answer this question was by making use of two theorems stated later in the same textbook. First, it is easy to prove that $|\mathbb{R}^\mathbb{R}-C|$ is an infinite set, since if it was finite we would have \begin{equation}2^{2^{\aleph_0}}=|\mathbb{R}^\mathbb{R}|=|\mathbb{R}^\mathbb{R}-C|+|C| \leq n+2^{\aleph_0}=2^{\aleph_0}< 2^{2^{\aleph_0}},\end{equation}which is a contradiction. Now I make use of a theorem requiring the axiom of choice: For every infinite set $S$ there exists a unique aleph $\aleph_\alpha$ such that $|S|=\aleph_\alpha$. So I let $|\mathbb{R}^\mathbb{R}-C|=\aleph_\alpha$ for some ordinal $\alpha$. Now assume $\aleph_\alpha < 2^{2^{\aleph_0}}$. Here I make use of another theorem, which i am not sure I am allowed to do: For every $\alpha$ and $\beta$ such that $\alpha \leq \beta$, we have $\aleph_\alpha+\aleph_\beta=\aleph_\beta$. So I basically then say either $|\mathbb{R}^\mathbb{R}-C|+|C|=\aleph_\alpha$ or $|\mathbb{R}^\mathbb{R}-C|+|C|=2^{\aleph_0}$, depending on whether $\aleph_\alpha \leq 2^{\aleph_0}$ or vice-versa. The assumption I am making is that $2^{\aleph_0}=\aleph_1$, which as I understand it is basically equivalent to the continuum hypothesis, as we are saying that the least uncountable number is $\aleph_1$ (but by the previous theorem requiring the axiom of choice this seems reasonable?). So anyway assuming I can do that, we again have a contradiction since then we are saying $|\mathbb{R}^\mathbb{R}|=2^{\aleph_0}<2^{2^{\aleph_0}}$ or $|\mathbb{R}^\mathbb{R}|=\aleph_\alpha < 2^{2^{\aleph_0}}$.
My main problem with my proof is that I am not making use of the result given in exercise 2.5 as suggested in the hint. Can anyone help with a proof making use of this result (stated in the second block quote above).